Solve $u_t+3uu_x=0$ , $u(x,0)=\left\{\begin{matrix} 2 & x<1\\ 0& x>1 \end{matrix}\right.$

Question

Solve the initial value problem $$u_t+3uu_x=0 ,\quad u(0,x)=\left\{\begin{matrix} 2 & x<1\\ 0& x>1 \end{matrix}\right.$$

My Idea:

given that $u_t+3uu_x=0$

Now $\frac{dt}{1}=\frac{dx}{3u}=\frac{du}{0}$

$\frac{dt}{1}=\frac{du}{0} \rightarrow u=c_1$

$\frac{dt}{1}=\frac{dx}{u} \Rightarrow \frac{dt}{1}=\frac{dx}{3c_1} \Rightarrow 3c_1t-x=c_2$

any one can help me from here

Answer 1

Let us assume that you want to solve $u_t + 3uu_x = 0$, with the proposed initial data $$ u(x,0) = \left\lbrace \begin{aligned} &2 &&\text{if}\quad x < 1 \, ,\\ &0 &&\text{if}\quad x > 1 \, .\\ \end{aligned} \right. $$ We follow the steps in this post. According to the Lax entropy condition, the solution to this Riemann problem is a shock wave (characteristic curves intersect in the $x$-$t$ plane), which speed $s$ is given by the Rankine-Hugoniot condition $$ s = \frac{3}{2}\left(0 + 2\right) = 3 \, . $$ One obtains the following solution: $$ u(x,t) = \left\lbrace \begin{aligned} &2 &&\text{if}\quad x < 1+st \, ,\\ &0 &&\text{if}\quad x > 1+st \, .\\ \end{aligned} \right. $$

Answer 2

Use the method of characteristics:

I assume you want to solve $$u_t+kuu_x=0.$$

Method of charactersitics results in: $$\dfrac{dt}{1}=\dfrac{dx}{ku}=\dfrac{du}{0}.$$

Using $$\dfrac{dt}{1}=\dfrac{du}{0}$$

gives $u=c_1$.

Using $$\dfrac{dt}{1}=\dfrac{dx}{ku}=\dfrac{dx}{kc_1}$$

$$\implies kc_1t=x+c_2 \implies c_2=kc_1t-x.$$

We know that

$$u=c_1=F(c_2)=F(kc_1t-x)=F(ktu-x)$$

in the last step I used $u=c_1$.